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BlackS17
@BlackS17
November 2021
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Решите неравенство очень срочно
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sedinalana
25*5^2x-29*5^x*2^x+4*2^2x≥0 /2^2x
25*(5/2)^2x-29*(5/2)^x+4≥0
(5/2)^x=a
25a²-29a+4≥0
D=841-400=441
a1=(29-21)/50=4/25
a2=(29+21)/50=1
a≤4/25⇒(5/2)^x≤(5/2)^-2⇒x≤-2
a≥1⇒(5/2)^x≥1⇒x≥0
x∈(-∞;-2] U [0;∞)
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Answers & Comments
25*(5/2)^2x-29*(5/2)^x+4≥0
(5/2)^x=a
25a²-29a+4≥0
D=841-400=441
a1=(29-21)/50=4/25
a2=(29+21)/50=1
a≤4/25⇒(5/2)^x≤(5/2)^-2⇒x≤-2
a≥1⇒(5/2)^x≥1⇒x≥0
x∈(-∞;-2] U [0;∞)